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Integration of the multiplication of normal cdf and exponential function

Definite integral of cdf of the form $Phileft(alpha+sqrt{d^2-frac{x^2}{2sigma^2}}right)$Unknown result in probability theory relating CDF of any density to the CDF of normal distributionhow to do integration $int_{-infty}^{+infty}exp(-x^n),mathrm{d}x$?How to compute normal integrals $int_{-infty}^inftyPhi(x)N(xmidmu,sigma^2),dx$ and $int_{-infty}^inftyPhi(x)N(xmidmu,sigma^2)x,dx$Integral involving CDF of a normal distributionDirect computation of $operatorname{log}(operatorname{cdf})$ for a normal distributionDifference between the Error function and Normal distribution function?Simplifying complicated integral including CDF of normal distributionDeriving the approximate moments of normal ratio distributionIs this the Cumulative Distribution Function for a Split Normal Distribution?

1

$begingroup$

I have to find the integral of
$$int_{M_0}^{infty} q(m, mu, sigma) beta e^{-beta(m-M_0)},mathrm{d}m,$$
where $q(m, mu, sigma)$ is the normal cumulative distribution function, $M_0$ is a constant, $m$ is the variable, and $beta$, $mu$, and $sigma$ are parameters. I have done the integration using the error function as follows:

begin{align}
int_{M_0}^{infty} q(m, mu, sigma) beta e^{-beta(m-M_0)},mathrm{d}m &=beta e^{beta M_0} int_{M_0}^{infty} frac{1}{2} Bigg[ 1+operatorname{erf}Bigg (frac{(m-mu)}{sigma sqrt(2)} Bigg ) Bigg] e^{-beta m },mathrm{d}m \
&=frac{1}{2} + frac{1}{2} beta e^{beta M_0} int_{M_0}^{infty} operatorname{erf} Bigg (frac{(m-mu)}{sigma sqrt(2)} Bigg)e^{- beta m},mathrm{d}m
end{align}

Here I get stuck. Could anyone please help solving this?

calculus integration normal-distribution numerical-calculus

share|cite|improve this question

edited Feb 23 at 20:16

Xander Henderson

14.9k103555

asked Feb 23 at 17:28

gultugultu

213

$endgroup$

add a comment | 

1

$begingroup$

I have to find the integral of
$$int_{M_0}^{infty} q(m, mu, sigma) beta e^{-beta(m-M_0)},mathrm{d}m,$$
where $q(m, mu, sigma)$ is the normal cumulative distribution function, $M_0$ is a constant, $m$ is the variable, and $beta$, $mu$, and $sigma$ are parameters. I have done the integration using the error function as follows:

begin{align}
int_{M_0}^{infty} q(m, mu, sigma) beta e^{-beta(m-M_0)},mathrm{d}m &=beta e^{beta M_0} int_{M_0}^{infty} frac{1}{2} Bigg[ 1+operatorname{erf}Bigg (frac{(m-mu)}{sigma sqrt(2)} Bigg ) Bigg] e^{-beta m },mathrm{d}m \
&=frac{1}{2} + frac{1}{2} beta e^{beta M_0} int_{M_0}^{infty} operatorname{erf} Bigg (frac{(m-mu)}{sigma sqrt(2)} Bigg)e^{- beta m},mathrm{d}m
end{align}

Here I get stuck. Could anyone please help solving this?

calculus integration normal-distribution numerical-calculus

share|cite|improve this question

edited Feb 23 at 20:16

Xander Henderson

14.9k103555

asked Feb 23 at 17:28

gultugultu

213

$endgroup$

add a comment | 

$begingroup$

I have to find the integral of
$$int_{M_0}^{infty} q(m, mu, sigma) beta e^{-beta(m-M_0)},mathrm{d}m,$$
where $q(m, mu, sigma)$ is the normal cumulative distribution function, $M_0$ is a constant, $m$ is the variable, and $beta$, $mu$, and $sigma$ are parameters. I have done the integration using the error function as follows:

begin{align}
int_{M_0}^{infty} q(m, mu, sigma) beta e^{-beta(m-M_0)},mathrm{d}m &=beta e^{beta M_0} int_{M_0}^{infty} frac{1}{2} Bigg[ 1+operatorname{erf}Bigg (frac{(m-mu)}{sigma sqrt(2)} Bigg ) Bigg] e^{-beta m },mathrm{d}m \
&=frac{1}{2} + frac{1}{2} beta e^{beta M_0} int_{M_0}^{infty} operatorname{erf} Bigg (frac{(m-mu)}{sigma sqrt(2)} Bigg)e^{- beta m},mathrm{d}m
end{align}

Here I get stuck. Could anyone please help solving this?

calculus integration normal-distribution numerical-calculus

share|cite|improve this question

edited Feb 23 at 20:16

Xander Henderson

14.9k103555

asked Feb 23 at 17:28

gultugultu

213

$endgroup$

share|cite|improve this question

edited Feb 23 at 20:16

Xander Henderson

14.9k103555

asked Feb 23 at 17:28

gultugultu

213

share|cite|improve this question

edited Feb 23 at 20:16

Xander Henderson

14.9k103555

asked Feb 23 at 17:28

gultugultu

213

edited Feb 23 at 20:16

Xander Henderson

14.9k103555

edited Feb 23 at 20:16

Xander Henderson

14.9k103555

asked Feb 23 at 17:28

gultugultu

213

asked Feb 23 at 17:28

gultugultu

213

2 Answers
2

active

oldest

votes

0

$begingroup$

I will present to you the solution in a sort of general step-by-step method. I hope you find it useful.

First of all, you have to use Partial Integration to get rid of the normal CDF function in the integral. Can you give it a go? If you have some difficulties, just ask me.

If you have done this correctly, you will end up with an integral of the form
$$ int_a^b e^{a x^2 + b x + c},mathrm{d} x $$
which you can't solve or can you? It seems difficult but it turns out that this is actually quite easy if you allow the normal CDF in your answer. The main idea in solving this integral is the same as splitting the square. So, this integral can be simplified to
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x $$
and this is then solved as
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x = e^{m} int_a^b e^{frac{(x + l)^2}{2(1/sqrt{2 k})^2}},mathrm{d} x = e^{m} cdot left[Phi(b) - Phi(a)right]cdot sqrt{2pi (1/sqrt{2k})^2} $$
where $Phi(z)$ is the CDF of a normal distribution with mean $-l$ and variance $1/(2k)$. This seems a bit random but all we did was actually writing the integrand to something which is similar to a normal pdf. Finally, note that we cannot simplify this any further in general. However, in your case, you have $b=infty$ and, thus, $Phi(b)=1$.

share|cite|improve this answer

answered Feb 23 at 19:29

Stan TendijckStan Tendijck

2,213415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Tendijick, please see my answer. I ended up there
  $endgroup$
  – gultu
  Feb 23 at 20:30
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not going to check tiny details but given that you want to know what is the next step is exactly what I explained in my answer. In the notation that I used you have $m=beta(sigma^2-2mu)/2$, $l=sigma^2beta-mu$ and $k=1/(2sigma^2)$ and $a=M_0$ and $b=infty$. Does that make sense?
  $endgroup$
  – Stan Tendijck
  Feb 23 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @yes, I follwoed your instructions
  $endgroup$
  – gultu
  Feb 23 at 20:59
  

  
  
  
  

add a comment | 

0

$begingroup$

the integral is:

$I= int_{M_0}^{infty} q(m,mu,sigma). beta e^{-beta (m-M_0)} dm \
=frac{1}{2}+frac{1}{2} beta e^{beta M_0} * I_{11}$

where,
$I_{11}= int_{M_0}^{infty} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta m} dm\
=erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm - int_{M_0}^{infty} { frac{d}{dm} erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm } dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0} + frac{1}{beta} frac{2}{sqrt{pi}} int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2}} e^{- beta m} dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0}+ frac{1}{beta} frac{2}{sqrt{pi}} I_{22}$

where,
$I_{22}= int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2 }} e^{- beta m} dm\
= int_{M_0}^{infty} e^{frac{{m-(mu-beta sigma^2) }^2}{2 sigma^2}+ frac{beta}{2} {sigma^2-2mu } } dm$

share|cite|improve this answer

edited Mar 11 at 11:46

answered Feb 23 at 20:30

gultugultu

213

$endgroup$

add a comment | 

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0

$begingroup$

I will present to you the solution in a sort of general step-by-step method. I hope you find it useful.

First of all, you have to use Partial Integration to get rid of the normal CDF function in the integral. Can you give it a go? If you have some difficulties, just ask me.

If you have done this correctly, you will end up with an integral of the form
$$ int_a^b e^{a x^2 + b x + c},mathrm{d} x $$
which you can't solve or can you? It seems difficult but it turns out that this is actually quite easy if you allow the normal CDF in your answer. The main idea in solving this integral is the same as splitting the square. So, this integral can be simplified to
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x $$
and this is then solved as
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x = e^{m} int_a^b e^{frac{(x + l)^2}{2(1/sqrt{2 k})^2}},mathrm{d} x = e^{m} cdot left[Phi(b) - Phi(a)right]cdot sqrt{2pi (1/sqrt{2k})^2} $$
where $Phi(z)$ is the CDF of a normal distribution with mean $-l$ and variance $1/(2k)$. This seems a bit random but all we did was actually writing the integrand to something which is similar to a normal pdf. Finally, note that we cannot simplify this any further in general. However, in your case, you have $b=infty$ and, thus, $Phi(b)=1$.

share|cite|improve this answer

answered Feb 23 at 19:29

Stan TendijckStan Tendijck

2,213415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Tendijick, please see my answer. I ended up there
  $endgroup$
  – gultu
  Feb 23 at 20:30
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not going to check tiny details but given that you want to know what is the next step is exactly what I explained in my answer. In the notation that I used you have $m=beta(sigma^2-2mu)/2$, $l=sigma^2beta-mu$ and $k=1/(2sigma^2)$ and $a=M_0$ and $b=infty$. Does that make sense?
  $endgroup$
  – Stan Tendijck
  Feb 23 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @yes, I follwoed your instructions
  $endgroup$
  – gultu
  Feb 23 at 20:59
  

  
  
  
  

add a comment | 

0

$begingroup$

I will present to you the solution in a sort of general step-by-step method. I hope you find it useful.

First of all, you have to use Partial Integration to get rid of the normal CDF function in the integral. Can you give it a go? If you have some difficulties, just ask me.

If you have done this correctly, you will end up with an integral of the form
$$ int_a^b e^{a x^2 + b x + c},mathrm{d} x $$
which you can't solve or can you? It seems difficult but it turns out that this is actually quite easy if you allow the normal CDF in your answer. The main idea in solving this integral is the same as splitting the square. So, this integral can be simplified to
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x $$
and this is then solved as
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x = e^{m} int_a^b e^{frac{(x + l)^2}{2(1/sqrt{2 k})^2}},mathrm{d} x = e^{m} cdot left[Phi(b) - Phi(a)right]cdot sqrt{2pi (1/sqrt{2k})^2} $$
where $Phi(z)$ is the CDF of a normal distribution with mean $-l$ and variance $1/(2k)$. This seems a bit random but all we did was actually writing the integrand to something which is similar to a normal pdf. Finally, note that we cannot simplify this any further in general. However, in your case, you have $b=infty$ and, thus, $Phi(b)=1$.

share|cite|improve this answer

answered Feb 23 at 19:29

Stan TendijckStan Tendijck

2,213415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Tendijick, please see my answer. I ended up there
  $endgroup$
  – gultu
  Feb 23 at 20:30
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not going to check tiny details but given that you want to know what is the next step is exactly what I explained in my answer. In the notation that I used you have $m=beta(sigma^2-2mu)/2$, $l=sigma^2beta-mu$ and $k=1/(2sigma^2)$ and $a=M_0$ and $b=infty$. Does that make sense?
  $endgroup$
  – Stan Tendijck
  Feb 23 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @yes, I follwoed your instructions
  $endgroup$
  – gultu
  Feb 23 at 20:59
  

  
  
  
  

add a comment | 

$begingroup$

I will present to you the solution in a sort of general step-by-step method. I hope you find it useful.

First of all, you have to use Partial Integration to get rid of the normal CDF function in the integral. Can you give it a go? If you have some difficulties, just ask me.

If you have done this correctly, you will end up with an integral of the form
$$ int_a^b e^{a x^2 + b x + c},mathrm{d} x $$
which you can't solve or can you? It seems difficult but it turns out that this is actually quite easy if you allow the normal CDF in your answer. The main idea in solving this integral is the same as splitting the square. So, this integral can be simplified to
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x $$
and this is then solved as
$$ int_a^b e^{k(x + l)^2 + m},mathrm{d} x = e^{m} int_a^b e^{frac{(x + l)^2}{2(1/sqrt{2 k})^2}},mathrm{d} x = e^{m} cdot left[Phi(b) - Phi(a)right]cdot sqrt{2pi (1/sqrt{2k})^2} $$
where $Phi(z)$ is the CDF of a normal distribution with mean $-l$ and variance $1/(2k)$. This seems a bit random but all we did was actually writing the integrand to something which is similar to a normal pdf. Finally, note that we cannot simplify this any further in general. However, in your case, you have $b=infty$ and, thus, $Phi(b)=1$.

share|cite|improve this answer

answered Feb 23 at 19:29

Stan TendijckStan Tendijck

2,213415

$endgroup$

share|cite|improve this answer

answered Feb 23 at 19:29

Stan TendijckStan Tendijck

2,213415

answered Feb 23 at 19:29

Stan TendijckStan Tendijck

2,213415

answered Feb 23 at 19:29

Stan TendijckStan Tendijck

2,213415

0

$begingroup$

the integral is:

$I= int_{M_0}^{infty} q(m,mu,sigma). beta e^{-beta (m-M_0)} dm \
=frac{1}{2}+frac{1}{2} beta e^{beta M_0} * I_{11}$

where,
$I_{11}= int_{M_0}^{infty} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta m} dm\
=erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm - int_{M_0}^{infty} { frac{d}{dm} erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm } dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0} + frac{1}{beta} frac{2}{sqrt{pi}} int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2}} e^{- beta m} dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0}+ frac{1}{beta} frac{2}{sqrt{pi}} I_{22}$

where,
$I_{22}= int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2 }} e^{- beta m} dm\
= int_{M_0}^{infty} e^{frac{{m-(mu-beta sigma^2) }^2}{2 sigma^2}+ frac{beta}{2} {sigma^2-2mu } } dm$

share|cite|improve this answer

edited Mar 11 at 11:46

answered Feb 23 at 20:30

gultugultu

213

$endgroup$

add a comment | 

0

$begingroup$

the integral is:

$I= int_{M_0}^{infty} q(m,mu,sigma). beta e^{-beta (m-M_0)} dm \
=frac{1}{2}+frac{1}{2} beta e^{beta M_0} * I_{11}$

where,
$I_{11}= int_{M_0}^{infty} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta m} dm\
=erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm - int_{M_0}^{infty} { frac{d}{dm} erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm } dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0} + frac{1}{beta} frac{2}{sqrt{pi}} int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2}} e^{- beta m} dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0}+ frac{1}{beta} frac{2}{sqrt{pi}} I_{22}$

where,
$I_{22}= int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2 }} e^{- beta m} dm\
= int_{M_0}^{infty} e^{frac{{m-(mu-beta sigma^2) }^2}{2 sigma^2}+ frac{beta}{2} {sigma^2-2mu } } dm$

share|cite|improve this answer

edited Mar 11 at 11:46

answered Feb 23 at 20:30

gultugultu

213

$endgroup$

add a comment | 

$begingroup$

the integral is:

$I= int_{M_0}^{infty} q(m,mu,sigma). beta e^{-beta (m-M_0)} dm \
=frac{1}{2}+frac{1}{2} beta e^{beta M_0} * I_{11}$

where,
$I_{11}= int_{M_0}^{infty} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta m} dm\
=erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm - int_{M_0}^{infty} { frac{d}{dm} erf(frac{(m-mu)}{sigma sqrt{2}}) int_{M_0}^{infty} e^{- beta m} dm } dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0} + frac{1}{beta} frac{2}{sqrt{pi}} int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2}} e^{- beta m} dm \
=frac{1}{beta} erf(frac{(m-mu)}{sigma sqrt{2}}) e^{- beta M_0}+ frac{1}{beta} frac{2}{sqrt{pi}} I_{22}$

where,
$I_{22}= int_{M_0}^{infty} e^{frac{(m-mu)^2}{2 sigma^2 }} e^{- beta m} dm\
= int_{M_0}^{infty} e^{frac{{m-(mu-beta sigma^2) }^2}{2 sigma^2}+ frac{beta}{2} {sigma^2-2mu } } dm$

share|cite|improve this answer

edited Mar 11 at 11:46

answered Feb 23 at 20:30

gultugultu

213

$endgroup$

share|cite|improve this answer

edited Mar 11 at 11:46

answered Feb 23 at 20:30

gultugultu

213

answered Feb 23 at 20:30

gultugultu

213

answered Feb 23 at 20:30

gultugultu

213